prove an operator is continuous

Question

In space $l^2$, consider an operator $R:l^2\to l^2$ defined by $Rx=y$ with $y_k=\frac{1}{k}x_k$ for all $k\geq1$. Prove the operator is continuous.

I know to show this operator is continuous, we need to show it is linear and bounded. But I don't know how to show it is bounded. I don't know how to calculate its bound.

Answer 1

We have $|y_k| = \frac{1}{k}|x_k| \leq |x_k|,$ hence also $||y||^2 = \sum_{k=1}^\infty y_k^2 \leq \sum_{k=1}^\infty x_k^2 = ||x||^2.$

So $||Rx|| \leq ||x||$ and in particular $||R||\leq 1$.

Answer 2

If $x=(x_k)$ has $\| x\| \leqslant +\infty$, as $\frac1{k^2}{x_k}^2 \leqslant {x_k}^2$, $\|Rx\| \leqslant \|x\|$, and $\|R\| \leqslant 1$. You can check that for $x=(1,0,0,\ldots)$, $\|Rx\| = \|x\|=1$ and thus, $\|R\|=1$.